Respuesta :

Answer:


Step-by-step explanation:

x decreases w/o bound, so x becomes a very large -ve no.

f(x)=5^(x-1) becomes 5^(large -ve no.) which will be almost zero


x increases w/o bound, so x becomes a very large +ve no.

f(x)=5^(x-1) becomes 5^(large +ve no.) which will keep increasing w/o bound


if u provide wat the choices are from ur Q in comments, i will pick the right one 4 u


wegnerkolmp2741o

Answer:

As x→-∞  f(x) →0

  As x→∞  f(x) →∞

Step-by-step explanation:

f(x) = 5^(x-1)

as x approaches negative infinity

f(x) = 5^ - ∞

     1/ 5^∞

   1/ big number

    0

As x→-∞  f(x) →0

as x approaches  infinity

f(x) = 5^  ∞

     5^∞

    big number

    As x→∞  f(x) →∞

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